There are many accounts of why A5 is not solvable on the internet. I'm recording my version here, because the proof involves certain ad-hoc choices which I want to make sure I can find off-hand in the future. We'll show that [A5, A5] = A5, thereby proving that A5 not solvable. This is useful for Galois theory, where we want to show tha A5 cannot be built as extensions of smaller cyclic groups.

§ Notation

I'll be using non-standard notation: (12);(34) means 'perform (12) then perform (34)'. I find this notation makes permutation composition intuitive for me. The ; is evocative of C-style languages, where we are ending a statement. I will be consistently using $[g, h] \equiv ghg^{-1}h^{-1}$ to denote the commutator.

§ permutations in A5

First, recall that A5 only has the even permutations in S5. So it can have zero, two, four, involutions that build it up. There can't be more after simplification, since S5 ony has 5 elements --- the largest sequence of transpositions we can do is (12)(23)(34)(45). So, in A5, we have:

• The identity permutation ().
• The transpositions (ij)(kl) where {i, j} and {k, l} do not overlap. From these, we get the 2-cycles.
• The transpositions (ij)(kl) where {i, j} and {k, l} overlap. Here we cannot have {i, j} = {k, l} since then we will just have a single transposition. So, let us assume that we have j = k. If we have any other equality, we can always flip the transpositions around to get to the normal form j = k:

(23);(12)= (32);(12) [(23) = (32)]= (32);(21) [(12) = (21)]

• In this case, we can show that such a transposition must be a cycle:

[a b c] -(32)->[a c b] -(21)->[c a b]

• Intuitively, we are pushing the element c backward, and allowing the other elements to take its place using the permutation (23);(12).

• So, from the transpositions of the form (ij)(kl) where {i, j} and {k, l} intersect, we get the 3-cycles.

• Finally, we can have the transpositions of the form (12)(23)(34)(45). It must be of this form, or some permutation of this form. Otherwise, we would have repeated elements, since these transpositions are packed "as close as possible". These generate the 5-cycles.

§ A5 is generated by 3-cycles.

We claim that we can write any element of $A5$ in terms of 3-cycles.

• The disjoint transpositions of the type (34)(12) can be written as (34)(23)(23)(12), because (23)(23) = e. This can be further broken down into ((34)(23)) ((23)(12)) which is two 2-cycles: (234); (123).

• The non-disjoint transpositions of the type (32)(21) are 3-cycles: (32)(21) = (123).

• 3-cycles are 3-cycles.

• Any 5-cycle an be written as two 3-cycles: (45)(34)(23)(12) can be written as ((45)(34))((23)(12)) which is two 3-cycles: (345); (123).

So, if we figure out how to write 3-cycles in terms of commutators, we win. Because the commutator subgroup of $A_n$ is generated by elements that can be written as $[g, h]$. If we can show that 3-cycles can be written as $[g, h]$, then every other element has a representation in terms of these 3-cycles, and are therefore elements of the commutator subgroup.

§ 3-cycles can be generated as commutators of 2-cycles:

• We saw how we can write a 3-cycle of the form C = (123) as (32)(21). We wish to write this as the commutator of two elements g, h: $C = [g, h]$.

• The idea is that we have the leftover elements 4, 5 that are unsused by C in A5[here is where 5 is important: 3 + 2 = 5, and we need two leftover elements ].

• We can use these two leftover elements 4, 5 to build elements g, hwhich cancel off, leaving us with (32)(21). We start with g = (32)___, h = (21)___ where the ___ is to be determined:

(32)___||(21)___||___(32)||___(21)  g        h        g^-1    h^-1

• It is important that g and h contain another tuple, because they are members of A5! We need them to be permutations having 2, 4, 6 transpositions.
• We insert (4 5) everywhere. These (4 5) can slide over the (2 1) and thereby harmlessly cancel:

(32)(45)||(21)(45)||(45)(32)||(45)(21)  g        h           g^-1       h^-1

• Simplify the above expression by moving the (45) over (21), (32):

(32)||(21)(45)(45)||(32)||(45)(45)(21)  g      h          g^-1    h^-1

• cancel the (45)(45) = e:

(32)||(21)||(32)||(21)  g   h    g^-1   h^-1

So we are left with (32);(21);(32);(21). This is the square of what we really wanted, C = (32);(21). However, since C is a 3-cycle, we know that $C = C^{-2}$. So, we can start with $C^{-1}$, use our trick to generate $C^{-2}$ which is equal to $C$. Since this works for any $C$, we have shown that we can generate 3-cycles from commutators of A5.

§ Alternate viewpoint on above proof

We have a 3-cycle s = (a b c). We first first a square root t such that t*t=s. To do this, we make t have the cycles of s spread out in gaps of 2:

t = (a _ _)t = (a _ b) [+2]t = (a c b) [+2, modulo]

It is hopefully clear that t*t = s:

t = (a c b)t*t: apply the cycle twice.t*t = a -(skip c) -> b      b -(skip a) -> c      c ->(skip b) -> a    = (a b c) = s

Now, we will write s = t*t and then find the commutator decomposition from it:

s = t*t  = (abc)(abc)  = (cb)(ba)(cb)(ba)  = (cb)|(ba)|(cb)|(ba)  = (cb)|(ba)|(cb)|(ba)     g     h   g-1   h-1

But there's a problem: this g and h do not belong to A5, they belong to S5. This is fixed by using a random (pq) which we know will exist .

§ Recap: How have we shown that A5 is not solvable?

what have we shown?

• 3-cycles can be written as $[g, h]$ for $g, h \in A_5$. Alternatively, we can say that 3-cycles belong to the commutator subgroup of $A_5$, since they can be written as commutators.
• any element in $A5$ can be written as the composition of 3-cycles.
• Hence, any element in $A5$ can be written as the composition of commutators.

In my mind, I think of it as:

arbitrary g= (3-cycle-1)(3-cycle-2)....(3-cycle-n)= [g, h][g2, h2]....[gn, hn]= member of [A5, A5]

Recall that $[A5, A5]$ is generated by commutators. It not only contains elements of the form $[g, h]$, but also all products of the form $[g, h][g', h']$. So we don't need to exhibit how to write a 5-cycle as some $[g, h]$. We just need to exhibit how to write as the product of commutators, which we have now shown.

§ Solvable implies simple

We can consider the other definition of simple. Let there be a chain of normal subgroups $G = N[0] \leq N[1] \leq N[1] \leq \dots \leq N[m] = e$, such that each quotient $N[i] / N[i+1]$ is abelian. Then, if $G$ is simple, this chain can only be $G = N[0] \leq N[1] = e$.

• If we want the quotient $G/N$ to be abelian, then we need the commutator subgroup $[G, G]$ to be a a subset of $N$.

• In our case, $[A_5, A_5] = A_5$. So if we want to remove the non-abelian-ness of A5, we need to quotient by the whole of $A5$.

• This means that any such chain will immediately collapse to $e$.

• So, it's impossible to build $A5$ using 'cycling components' starting from $\{e\}$. Viewed from the field theoretic perspective, this means that it's impossible to reach a polynomial whose splitting field has galois group A5 by simply appending cycles.

§ Nagging doubt: Did we depend on our numbering of cycles?

In all my proofs, I had used one 3-cycle, or 5-cycle, or 2-cycle to argue that it all works out. Is this really legal? Perhaps the argument written for the 3-cycle C = (123) will break down for D = (321). Fear not!

• We will show that all 3-cycles are conjugate to each other. So, we can always relabel a 3-cycle within A5.
• It is easy to note that $g[k, l]g^{-1} = [gkg^{-1}, glg^{-1}]$. This shows that the commutator subgroup is closed under conjugation. It better be, because it ought to be normal for us to take quotients from it.
• Combining these facts, if we show that (123) is in [A5, A5], then some other cycle (ijk) can be conjugated to (123). Since the commutator subgroup is closed under conjugation, we have that (ijk) is a member of [A5, A5].

§ All 3-cycles are conjugate to each other in A5.

• Given two 3-cycles C=(abc) and D=(pqr), at least one of a, b, c must be equal to one of p, q, r. Since each a, b, c is unique, and each p, q, r is unique, for them to not overlap, we would need 6 elements. But we only have 5, so there must be some overlap:

a   b   c1 2 3 4 5  p   q r

So, we will perform our proof assuming there is 1 overlap, 2 overlap, 3 overlap. Recall that if C = (a b c) is a cycle and s is a permutation, then the action of conjugating C with s produces a permutation (s(a) s(b) s(c)). We will prove our results by finding an s, and then making s even . This is the difficult part of the proof, since we need to show that all 3-cycles are conjugate in A5 . We will write s as two distinct transpositions, which will guarantee that it belongs to A5.

• Case 1: (abx) and (pqx) have a single element x in common:

C = (abx)D = (pqx)s: send a to p, b to qs = (ap)(bq)C = (abx) -conj s-> (pqx) = D

• Case 2: (axy) and (pxy) have two elements in common, x and y. Naively, we would pick s: send x to y. But this is odd, so this isn't a member of A5. To make it even, we rearrange D = (pxy) as D = (yxp). This lets us go from C to D by relabelling a to y, y to p. This permutation is even since it has two distinct transpositions.

C = (axy)D = (pxy) = (yxp) [cyclic property]s: send a to y, y to ps = (ay)(yp)C = (axy) -conj s-> (yxp) = D

• Case 3: (xyz) and (xyz) have all three elements in common, x, y, z. Here we can conjugate by identity and we are done.

§ Why do we care about solvable?

• Roughly, we can look at the solvability criterion as giving us a way to build our group $G$ from a series of extensions $N[1], N[2], \dots$. This extension is special, because at each step, we are adding a cyclic group.

• When we want to write a solution using nth roots, we can only add the nth roots of unity, a "cyclic" component. So, any element we can reach by using nth roots ought to be able to be written down as an extension of cyclic elements.

§ SAGE code to play around with commutators of A5:

• Create a dictionary m which maps each element of A5 to the commutators that create it.

from collections import defaultdictm = defaultdict(set)A5 = AlternatingGroup(5)S5 = SymmetricGroup(5) # if necessaryfor g in A5:    for h in A5:        m[g * h * g^(-1) * h^(-1)] |= { (g, h) }# all 60 elem can be written in terms of commutatorsprint("number of elem generated as commutator: " + str(len(m.keys())))# Show how to access elements of A5 and their commutator representationcyc5 = A5("(1, 2, 3, 4, 5)")cyc3 = A5("(1, 2, 3)")cyc2disj = A5("(1, 2) (3, 4)")print(m[cyc5])print(m[cyc3])print(m[cyc2disj])

§ Writing each element in A5 directly as a commutator

We have shown how to write 3-cycles as the commutator of 2-cycles. We will now show how to do this for disjoint 2-cycles and 5-cycles as a matter of enlightenment.

§ Writing disjoint 2-cycles as commutator

First, we will write a two disjoint two cycles as the square root of a 4-cycle. We will then show how to write this 4-cycle as two 3-cycles.

s = (12)(34)

Note that if we choose t = (abcd), then t*t will exchange the first and third elements a <-> c, and the second and fourth elements b <-> d. So, if we choose:

t = (1324)t*t = (12) (34)

Next, we need to write this t*t as [g, h] for g, h from A5.

t*t = (1324)(1324)    = (42)(23)(31);(42)(23)(31)    = (42)(23)(31);(42)(23)(31)    = (42)(23)(31);(23)(23);(42)(23)(31)                   ^^^^^^^^ inserted    = (42)(23)|(31)(23)|(23)(42)|(23)(31)          g   |    h   |   g'   |   h'    = [(42)(23), (31)(23)]

Where both (42)(23), and (31)(23) are members of A5.

§ Writing 3-cycle as commutator

In the description of showing how to generate 3-cycles, we do this explicitly.

§ Writing 5-cycle as commutator

Let s = (1 2 3 4 5). we once again find a square root of s. To build this, we will build an element with the elements of s written with gaps of 2:

t = (1 _ _ _ _)  = (1 _ 2 _ _)  [+2 index]  = (1 _ 2 _ 3)  [+2 index, wrap]  = (1 4 2 _ 3)  [+2 index, wrap]  = (1 4 2 5 3)  [+2 index, wrap]

It should be clear how t*t = s: When we take s = t*t, the resulting permutation s will move an element j = t[i] to k = t[i+2]. But we have built t such that t[i+2] = s[i+1]. So we will move the element according to how s pleases:

t = (1 4 2 5 3)t*t = 1 -> (4 skip) -> 2      2 -> (5 skip) -> 3      3 -> (1 skip) -> 4      3 -> (2 skip) -> 5      5 -> (3 skip) -> 1t*t = (1 2 3 4 5) = s

We will now use t*t to write the commutator:

s = t*t  = (35)(52)(24)(41);(35)(52)(24)(41)  =  =  =  = (1, 2)(3, 5)|(1, 5)(2, 4)|(3, 5)(1, 2)|(2, 4)(1, 5)  = (1, 2)(3, 5)|(1, 5)(2, 4)|(3, 5)(1, 2)|(2, 4)(1, 5)         g             h          g^{-1}       h^{-1}

§ To think: relationship between square roots and commutators?